diff --git a/solution/2600-2699/2643.Row With Maximum Ones/README.md b/solution/2600-2699/2643.Row With Maximum Ones/README.md index d29d5170fab3a..1d6d883d7d79f 100644 --- a/solution/2600-2699/2643.Row With Maximum Ones/README.md +++ b/solution/2600-2699/2643.Row With Maximum Ones/README.md @@ -32,7 +32,7 @@ tags: 
 输入：mat = [[0,1],[1,0]]
 输出：[0,1]
-解释：两行中 1 的数量相同。所以返回下标最小的行，下标为 0 。该行 1 的数量为 1 。所以，答案为 [0,1] 。 
+解释：两行中 1 的数量相同。所以返回下标最小的行，下标为 0 。该行 1 的数量为 1 。所以，答案为 [0,1] 。

示例 2：

@@ -69,9 +69,16 @@ tags: ### 方法一：模拟 -我们直接遍历矩阵，统计每一行中 $1$ 的个数，更新最大值和对应的行下标。注意，如果当前行的 $1$ 的个数与最大值相等，我们需要选择行下标较小的那一行。 +我们初始化一个数组 $\textit{ans} = [0, 0]$，用于记录最多 $1$ 的行的下标和 $1$ 的数量。 -时间复杂度 $(m \times n)$，其中 $m$ 和 $n$ 分别为矩阵的行数和列数。空间复杂度 $O(1)$。 +然后遍历矩阵的每一行，对于每一行： + +- 计算该行 $1$ 的数量 $\textit{cnt}$（由于矩阵中只包含 $0$ 和 $1$，我们可以直接对该行求和）； +- 如果 $\textit{ans}[1] < \textit{cnt}$，则更新 $\textit{ans} = [i, \textit{cnt}]$。 + +遍历结束后，返回 $\textit{ans}$。 + +时间复杂度 $O(m \times n)$，其中 $m$ 和 $n$ 分别是矩阵的行数和列数。空间复杂度 $O(1)$。 @@ -82,7 +89,7 @@ class Solution: def rowAndMaximumOnes(self, mat: List[List[int]]) -> List[int]: ans = [0, 0] for i, row in enumerate(mat): - cnt = row.count(1) + cnt = sum(row) if ans[1] < cnt: ans = [i, cnt] return ans @@ -97,9 +104,7 @@ class Solution { for (int i = 0; i < mat.length; ++i) { int cnt = 0; for (int x : mat[i]) { - if (x == 1) { - ++cnt; - } + cnt += x; } if (ans[1] < cnt) { ans[0] = i; @@ -119,13 +124,9 @@ public: vector rowAndMaximumOnes(vector>& mat) { vector ans(2); for (int i = 0; i < mat.size(); ++i) { - int cnt = 0; - for (auto& x : mat[i]) { - cnt += x == 1; - } + int cnt = accumulate(mat[i].begin(), mat[i].end(), 0); if (ans[1] < cnt) { - ans[0] = i; - ans[1] = cnt; + ans = {i, cnt}; } } return ans; @@ -137,16 +138,14 @@ public: ```go func rowAndMaximumOnes(mat [][]int) []int { - ans := make([]int, 2) + ans := []int{0, 0} for i, row := range mat { cnt := 0 for _, x := range row { - if x == 1 { - cnt++ - } + cnt += x } if ans[1] < cnt { - ans[0], ans[1] = i, cnt + ans = []int{i, cnt} } } return ans @@ -158,8 +157,8 @@ func rowAndMaximumOnes(mat [][]int) []int { ```ts function rowAndMaximumOnes(mat: number[][]): number[] { const ans: number[] = [0, 0]; - for (let i = 0; i < mat.length; ++i) { - const cnt = mat[i].reduce((a, b) => a + b); + for (let i = 0; i < mat.length; i++) { + const cnt = mat[i].reduce((sum, num) => sum + num, 0); if (ans[1] < cnt) { ans[0] = i; ans[1] = cnt; @@ -175,20 +174,34 @@ function rowAndMaximumOnes(mat: number[][]): number[] { impl Solution { pub fn row_and_maximum_ones(mat: Vec>) -> Vec { let mut ans = vec![0, 0]; - for (i, row) in mat.iter().enumerate() { - let cnt = row.iter().filter(|&v| *v == 1).count() as i32; + let cnt = row.iter().sum(); if ans[1] < cnt { - ans[0] = i as i32; - ans[1] = cnt; + ans = vec![i as i32, cnt]; } } - ans } } ``` +#### C# + +```cs +public class Solution { + public int[] RowAndMaximumOnes(int[][] mat) { + int[] ans = new int[2]; + for (int i = 0; i < mat.Length; i++) { + int cnt = mat[i].Sum(); + if (ans[1] < cnt) { + ans = new int[] { i, cnt }; + } + } + return ans; + } +} +``` + diff --git a/solution/2600-2699/2643.Row With Maximum Ones/README_EN.md b/solution/2600-2699/2643.Row With Maximum Ones/README_EN.md index 236220b1491cd..f6b88dcb4e3eb 100644 --- a/solution/2600-2699/2643.Row With Maximum Ones/README_EN.md +++ b/solution/2600-2699/2643.Row With Maximum Ones/README_EN.md @@ -31,7 +31,7 @@ tags: 
 Input: mat = [[0,1],[1,0]]
 Output: [0,1]
-Explanation: Both rows have the same number of 1's. So we return the index of the smaller row, 0, and the maximum count of ones (1). So, the answer is [0,1]. 
+Explanation: Both rows have the same number of 1's. So we return the index of the smaller row, 0, and the maximum count of ones (1). So, the answer is [0,1].

Example 2:

@@ -68,9 +68,16 @@ tags: ### Solution 1: Simulation -We directly traverse the matrix, count the number of $1$s in each row, and update the maximum value and the corresponding row index. Note that if the number of $1$s in the current row is equal to the maximum value, we need to choose the row with the smaller index. +We initialize an array $\textit{ans} = [0, 0]$ to store the index of the row with the most $1$s and the count of $1$s. -The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively. The space complexity is $O(1)$. +Then, we iterate through each row of the matrix: + +- Compute the number of $1$s in the current row, denoted as $\textit{cnt}$ (since the matrix contains only $0$s and $1$s, we can directly sum up the row). +- If $\textit{ans}[1] < \textit{cnt}$, update $\textit{ans} = [i, \textit{cnt}]$. + +After finishing the iteration, we return $\textit{ans}$. + +The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns in the matrix, respectively. The space complexity is $O(1)$. @@ -81,7 +88,7 @@ class Solution: def rowAndMaximumOnes(self, mat: List[List[int]]) -> List[int]: ans = [0, 0] for i, row in enumerate(mat): - cnt = row.count(1) + cnt = sum(row) if ans[1] < cnt: ans = [i, cnt] return ans @@ -96,9 +103,7 @@ class Solution { for (int i = 0; i < mat.length; ++i) { int cnt = 0; for (int x : mat[i]) { - if (x == 1) { - ++cnt; - } + cnt += x; } if (ans[1] < cnt) { ans[0] = i; @@ -118,13 +123,9 @@ public: vector rowAndMaximumOnes(vector>& mat) { vector ans(2); for (int i = 0; i < mat.size(); ++i) { - int cnt = 0; - for (auto& x : mat[i]) { - cnt += x == 1; - } + int cnt = accumulate(mat[i].begin(), mat[i].end(), 0); if (ans[1] < cnt) { - ans[0] = i; - ans[1] = cnt; + ans = {i, cnt}; } } return ans; @@ -136,16 +137,14 @@ public: ```go func rowAndMaximumOnes(mat [][]int) []int { - ans := make([]int, 2) + ans := []int{0, 0} for i, row := range mat { cnt := 0 for _, x := range row { - if x == 1 { - cnt++ - } + cnt += x } if ans[1] < cnt { - ans[0], ans[1] = i, cnt + ans = []int{i, cnt} } } return ans @@ -157,8 +156,8 @@ func rowAndMaximumOnes(mat [][]int) []int { ```ts function rowAndMaximumOnes(mat: number[][]): number[] { const ans: number[] = [0, 0]; - for (let i = 0; i < mat.length; ++i) { - const cnt = mat[i].reduce((a, b) => a + b); + for (let i = 0; i < mat.length; i++) { + const cnt = mat[i].reduce((sum, num) => sum + num, 0); if (ans[1] < cnt) { ans[0] = i; ans[1] = cnt; @@ -174,20 +173,34 @@ function rowAndMaximumOnes(mat: number[][]): number[] { impl Solution { pub fn row_and_maximum_ones(mat: Vec>) -> Vec { let mut ans = vec![0, 0]; - for (i, row) in mat.iter().enumerate() { - let cnt = row.iter().filter(|&v| *v == 1).count() as i32; + let cnt = row.iter().sum(); if ans[1] < cnt { - ans[0] = i as i32; - ans[1] = cnt; + ans = vec![i as i32, cnt]; } } - ans } } ``` +#### C# + +```cs +public class Solution { + public int[] RowAndMaximumOnes(int[][] mat) { + int[] ans = new int[2]; + for (int i = 0; i < mat.Length; i++) { + int cnt = mat[i].Sum(); + if (ans[1] < cnt) { + ans = new int[] { i, cnt }; + } + } + return ans; + } +} +``` + diff --git a/solution/2600-2699/2643.Row With Maximum Ones/Solution.cpp b/solution/2600-2699/2643.Row With Maximum Ones/Solution.cpp index 79bcbe90b4e44..5df58522c7b0a 100644 --- a/solution/2600-2699/2643.Row With Maximum Ones/Solution.cpp +++ b/solution/2600-2699/2643.Row With Maximum Ones/Solution.cpp @@ -3,15 +3,11 @@ class Solution { vector rowAndMaximumOnes(vector>& mat) { vector ans(2); for (int i = 0; i < mat.size(); ++i) { - int cnt = 0; - for (auto& x : mat[i]) { - cnt += x == 1; - } + int cnt = accumulate(mat[i].begin(), mat[i].end(), 0); if (ans[1] < cnt) { - ans[0] = i; - ans[1] = cnt; + ans = {i, cnt}; } } return ans; } -}; \ No newline at end of file +}; diff --git a/solution/2600-2699/2643.Row With Maximum Ones/Solution.cs b/solution/2600-2699/2643.Row With Maximum Ones/Solution.cs new file mode 100644 index 0000000000000..1b3b9999f632a --- /dev/null +++ b/solution/2600-2699/2643.Row With Maximum Ones/Solution.cs @@ -0,0 +1,12 @@ +public class Solution { + public int[] RowAndMaximumOnes(int[][] mat) { + int[] ans = new int[2]; + for (int i = 0; i < mat.Length; i++) { + int cnt = mat[i].Sum(); + if (ans[1] < cnt) { + ans = new int[] { i, cnt }; + } + } + return ans; + } +} diff --git a/solution/2600-2699/2643.Row With Maximum Ones/Solution.go b/solution/2600-2699/2643.Row With Maximum Ones/Solution.go index e3b9d532c6c55..8951c9e9b1aa4 100644 --- a/solution/2600-2699/2643.Row With Maximum Ones/Solution.go +++ b/solution/2600-2699/2643.Row With Maximum Ones/Solution.go @@ -1,15 +1,13 @@ func rowAndMaximumOnes(mat [][]int) []int { - ans := make([]int, 2) + ans := []int{0, 0} for i, row := range mat { cnt := 0 for _, x := range row { - if x == 1 { - cnt++ - } + cnt += x } if ans[1] < cnt { - ans[0], ans[1] = i, cnt + ans = []int{i, cnt} } } return ans -} \ No newline at end of file +} diff --git a/solution/2600-2699/2643.Row With Maximum Ones/Solution.java b/solution/2600-2699/2643.Row With Maximum Ones/Solution.java index 1658ecd1f784d..0c3880cb7020a 100644 --- a/solution/2600-2699/2643.Row With Maximum Ones/Solution.java +++ b/solution/2600-2699/2643.Row With Maximum Ones/Solution.java @@ -4,9 +4,7 @@ public int[] rowAndMaximumOnes(int[][] mat) { for (int i = 0; i < mat.length; ++i) { int cnt = 0; for (int x : mat[i]) { - if (x == 1) { - ++cnt; - } + cnt += x; } if (ans[1] < cnt) { ans[0] = i; @@ -15,4 +13,4 @@ public int[] rowAndMaximumOnes(int[][] mat) { } return ans; } -} \ No newline at end of file +} diff --git a/solution/2600-2699/2643.Row With Maximum Ones/Solution.py b/solution/2600-2699/2643.Row With Maximum Ones/Solution.py index af13862cc0378..860faeeac4ce2 100644 --- a/solution/2600-2699/2643.Row With Maximum Ones/Solution.py +++ b/solution/2600-2699/2643.Row With Maximum Ones/Solution.py @@ -2,7 +2,7 @@ class Solution: def rowAndMaximumOnes(self, mat: List[List[int]]) -> List[int]: ans = [0, 0] for i, row in enumerate(mat): - cnt = row.count(1) + cnt = sum(row) if ans[1] < cnt: ans = [i, cnt] return ans diff --git a/solution/2600-2699/2643.Row With Maximum Ones/Solution.rs b/solution/2600-2699/2643.Row With Maximum Ones/Solution.rs index da124bf28e74f..9f88929a8d6fa 100644 --- a/solution/2600-2699/2643.Row With Maximum Ones/Solution.rs +++ b/solution/2600-2699/2643.Row With Maximum Ones/Solution.rs @@ -1,15 +1,12 @@ impl Solution { pub fn row_and_maximum_ones(mat: Vec>) -> Vec { let mut ans = vec![0, 0]; - for (i, row) in mat.iter().enumerate() { - let cnt = row.iter().filter(|&v| *v == 1).count() as i32; + let cnt = row.iter().sum(); if ans[1] < cnt { - ans[0] = i as i32; - ans[1] = cnt; + ans = vec![i as i32, cnt]; } } - ans } } diff --git a/solution/2600-2699/2643.Row With Maximum Ones/Solution.ts b/solution/2600-2699/2643.Row With Maximum Ones/Solution.ts index 879c1afd9b73a..2d8f1513c806e 100644 --- a/solution/2600-2699/2643.Row With Maximum Ones/Solution.ts +++ b/solution/2600-2699/2643.Row With Maximum Ones/Solution.ts @@ -1,7 +1,7 @@ function rowAndMaximumOnes(mat: number[][]): number[] { const ans: number[] = [0, 0]; - for (let i = 0; i < mat.length; ++i) { - const cnt = mat[i].reduce((a, b) => a + b); + for (let i = 0; i < mat.length; i++) { + const cnt = mat[i].reduce((sum, num) => sum + num, 0); if (ans[1] < cnt) { ans[0] = i; ans[1] = cnt;